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1 year ago

15

Which reactant is unlikely to produce the indicated product upon strong heating?

1 answer:

Ilya [14]1 year ago

8 0

2-Methyl-4-oxo-pentanoic acid is unlikely to produce 2-Methyl-3-butanone upon strong heating.

Upon heating, the β ketoacid becomes unstable and decarboxylates, leading to the formation of the methyl ketone.

A carboxylic acid is an organic acid that contains a carboxyl group (C(=O)OH) attached to an R-group. The general formula of a carboxylic acid is R−COOH or R−CO2H, with R referring to the alkyl, alkenyl, aryl, or other group.

Carboxylic acids occur widely. Important examples include the amino acids and fatty acids. Deprotonation of a carboxylic acid gives a carboxylate anion.

Full question :

Q. Which reactant is unlikely to produce the indicated product upon strong heating?

A) 2,2-Dimethylpropanedioic acid 2-methylpropanoic acid
B) 2-Ethylpropanedioic acid Butanoic acid
C) 2-Methyl-3-oxo-pentanoic acid 3-Pentanone
D) 2-Methyl-4-oxo-pentanoic acid 2-Methyl-3-butanone
E) 4-Methyl-3-oxo-heptanoic acid 3-Methyl-2-hexanone

Hence, option (D) is correct.

Learn more about carboxylic acid here : brainly.com/question/26855500

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The electronegativity of nitrogen (n) is 3.0, while the electronegativity of hydrogen (h) is 2.1. knowing this, consider how the

Goryan [66]

The electronegativity of nitrogen (N) is 3.0, while the electronegativity of hydrogen (H) is 2.1. As it can be seen that nitrogen (N) is more electronegative than that of hydrogen (H),

So electron pairs are attracted towards nitrogen and thus it carries a partial negative charge and hydrogen carries a partial positive charge. The image of electron distribution is attached as follows.

Thus NH₃ is a polar molecule .

8 0

3 years ago

What is the pH of a 3.9•10-8 M OH- solution

Mice21 [21]

POH will be -log[conc of OH]
-log (3.9E-08) = 7.409
pH = 14- pOH
pH = 14 - 7.409
pH = 6.59

3 0

3 years ago

For the titration of 25.00 mL of 0.150 M HCl with 0.250 M NaOH, calculate:

Leviafan [203]

1) Chemical reaction

HCl        +       NaOH      --->      NaCl + H2O

25.0 ml
0.150 M            0.250M

2) 50% completion => 0.025 l * 0.150 M * (1/2) = 0.001875 mol HCl consumed and 0.001875 mol HCl in solution

0.001875 mol HCl => 0.001875 mol H(+)

Volume = Volume of HCl solution + Volumen of NaOH solution added

Volume of HCl solution = 0.0250 l

Volume of NaOH = n / M = 0.001875 mol / 0.250M = 0.0075 l

Total volume = 0.0250 l + 0.0075 l = 0.0325 l

[H+] = 0.001875 mol / 0.0325 l = 0.05769 M

pH = - log [H+] = - log (0.05769) = 1.23

Answer: 1.23

3) Equivalence point

0.02500 l * 0.150 M = 0.250M * V

=> V = 0.02500 * 0.150 / 0.250 = 0.015 l

4) 1.00 ml NaOH added beyond the equivalence point

1.00 ml * 1 l / 1000 ml * 0.250 M = 0.00025 mol NaOH in excess

0.00025 mol NaOH = 0.00025 mol OH-

Volume of the solution = 0.02500 l + 0.015 l + 1.00/1000 l = 0.041 l

[OH-] = 0.00025 mol / 0.041 l = 0.00610 M

pOH = - log (0.00610) = 2.21

pH + pOH = 14 => pH = 14 - pOH = 14 - 2.21 = 11.76

Answer: 11.76

6 0

3 years ago

What mass of copper sulphate dissolves in 100g of water at 70 °C

ValentinkaMS [17]

The answer mass of copper sulphate is

46g

6 0

3 years ago

The Haber Process synthesizes ammonia at elevated temperatures and pressures. Suppose you combine 1580 L of nitrogen gas and 351

ikadub [295]

Answer : The volume of reactant measured at STP left over is 409.9 L

Explanation :

First we have to calculate the moles of $N_2$ and $H_2$ by using ideal gas equation.

<u>For $N_2$ :</u>

$PV_{N_2}=n_{N_2}RT$

where,

P = Pressure of gas at STP = 1 atm

V = Volume of $N_2$ gas = 1580 L

n = number of moles $N_2$ = ?

R = Gas constant = $0.0821L.atm/mol.K$

T = Temperature of gas at STP = 273 K

Putting values in above equation, we get:

$1atm\times 1580L=n_{N_2}\times (0.0821L.atm/mol.K)\times 273K$

$n_{N_2}=70.49mole$

<u>For $H_2$ :</u>

$PV_{H_2}=n_{H_2}RT$

where,

P = Pressure of gas at STP = 1 atm

V = Volume of $H_2$ gas = 3510 L

n = number of moles $H_2$ = ?

R = Gas constant = $0.0821L.atm/mol.K$

T = Temperature of gas at STP = 273 K

Putting values in above equation, we get:

$1atm\times 3510L=n_{H_2}\times (0.0821L.atm/mol.K)\times 273K$

$n_{H_2}=156.6mole$

Now we have to calculate the limiting and excess reagent.

The balanced chemical reaction is,

$N_2(g)+3H_2(g)\rightarrow 2NH_3(g)$

From the balanced reaction we conclude that

As, 3 mole of $H_2$ react with 1 mole of $N_2$

So, 156.6 moles of $H_2$ react with $\frac{156.6}{3}\times 1=52.2$ moles of $N_2$

From this we conclude that, $N_2$ is an excess reagent because the given moles are greater than the required moles and $H_2$ is a limiting reagent and it limits the formation of product.

Now we have to calculate the excess moles of $N_2$ reactant (unreacted gas).

Excess moles of $N_2$ reactant = 70.49 - 52.2 = 18.29 moles

Now we have to calculate the volume of reactant, measured at STP, is left over.

$PV=nRT$

where,

P = Pressure of gas at STP = 1 atm

V = Volume of gas = ?

n = number of moles of unreacted gas = 18.29 moles

R = Gas constant = $0.0821L.atm/mol.K$

T = Temperature of gas at STP = 273 K

Putting values in above equation, we get:

$1atm\times V=18.29mole\times (0.0821L.atm/mol.K)\times 273K$

$V=409.9L$

Therefore, the volume of reactant measured at STP left over is 409.9 L

8 0

3 years ago